A simple {{w|Intercept theorem|formula}} can be used to find the size on earth of a celestial object when the size of and distance to the object is known. This is done by taking the radius of the earth, multiplying by the diameter of the object, and dividing by the distance to the object from the center of the earth.

A simple {{w|Intercept theorem|formula}} can be used to find the size on earth of a celestial object when the size of and distance to the object is known. This is done by taking the radius of the earth, multiplying by the diameter of the object, and dividing by the distance to the object from the center of the earth.

* The {{w|Sun}} and the {{w|Moon}} have approximately the same size (around 0.5 degrees of arc) when seen from the Earth.

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* {{w|Mercury (planet)|Mercury}}, {{w|Venus}}, {{w|Mars}}, {{w|Jupiter}}, {{w|Saturn}}, {{w|Uranus}}, and {{w|Neptune}} are the other planets of the {{w|Solar System}}.

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* {{w|Io (moon)|Io}}, {{w|Europa (moon)|Europa}}, {{w|Ganymede}}, and {{w|Calisto}} are the main moons of Jupiter; {{w|Titan (moon)|Titan}} is the largest moon of Saturn; and {{w|Triton (moon)|Triton}} is the largest moon of Neptune. {{w|Ceres (dwarf planet)|Ceres}} and {{w|Pluto (dwarf planet)|Pluto}} are {{w|dwarf planet}}s.

Revision as of 12:54, 11 October 2013

Title text: If the celestial sphere were mapped to the Earth's surface, astronomy would get a LOT easier; you'd just need a magnifying glass.

Explanation

This explanation may be incomplete or incorrect:Please include the reason why this explanation is incomplete, like this: {{incomplete|reason}}If you can address this issue, please edit the page! Thanks.

This comic is a comparison of the sizes of celestial objects when they are measured on earth's surface as shown under the title, and the sizes of everyday objects actually on earth. London's M25 motorway is around 60 km (35 miles) across, a soccer field is about 100 meters long, a ping pong table is 274 centimeters long, a laptop is about 35 centimeters across (this one resembles an Apple MacBook Pro, the tilde symbol on a keyboard is about 5 millimeters long, and a cell of E. coli is about 2 micrometers long.

A simple formula can be used to find the size on earth of a celestial object when the size of and distance to the object is known. This is done by taking the radius of the earth, multiplying by the diameter of the object, and dividing by the distance to the object from the center of the earth.

The astronomical objects referenced in the panels are:

The Sun and the Moon have approximately the same size (around 0.5 degrees of arc) when seen from the Earth.

Transcript

Discussion

What is the meaning of "football field" in panel #2? --Kevang (talk) 04:50, 11 October 2013 (UTC)

I was wondering the same thing. Probably misplaced text. Irino. (talk) 05:49, 11 October 2013 (UTC)

It does seem to be misplaced, but if that's the only glitch, this is the only panel without a unique reference object. "20 football pitches long" isn't all that easy to grasp. jameslucas(" " / +) 09:09, 11 October 2013 (UTC)

The image is fixed by Randall. I did an update here.--Dgbrt (talk) 11:28, 11 October 2013 (UTC)

I haven't done any lookups or maths to check these, but give the size of these as "stars" in the sky, everything from panel 2 onwards seems to me to be an order of magnitude or two too large. Mark Hurd (talk) 05:17, 11 October 2013 (UTC)

Not really. You see the stars and planets as points because their angular size is lower than your eyes' resolution. They have measurable (or, in case of really distant or small objects, computable) angular sizes. For stars etc. these angular sizes are really small - but Earth is quite big, so if you cut a portion of a sphere the radius of Earth corresponding to these small solid angles, you get sizable areas. I haven't checked Randall's math, but I'd rather believe his results. If it is non-intuitive for you consider the Sun and Moon example - when observed by naked eye, the Moon looks for you as being the size of a dime held up in your hand - and yet it's shadow during an eclipse covers quite an area of Earth's surface. It is true that sizes of some of these "footprints" are quite surprising compared to other ones. 89.174.214.74 08:55, 11 October 2013 (UTC)

Definitely surprising. I'll put faith in Randall doing his math correctly, but still needed to check on a couple of these because they did elicit a "What? No. Really? Can't be." reaction. Using the formula described in the Explanation above, for Venus I get 12742 km (Earth radius) * 12104 km (Venus diameter) / 38000000 (shortest distance to Venus) = 2.03 km.

Hard to picture that something that is such a small dot in the sky is actually directly over such a large patch of ground. But there you are.

The question that sprung to my mind was, which distance is he using for the planets and asteroids, since those vary hugely depending on where objects are relative to each other along their orbits. Is he going with closest approach, maybe? Or the distance that we happen to be at just this instant? --Rmharman (talk) 21:42, 11 October 2013 (UTC)

I just checked for Deimos, and got to ~50 mio km, so that´d be the closest approach. --Wilberforce (talk) 13:24, 13 October 2013 (UTC)

Does someone know how to use LaTeX formulas? And if so, can they translate my formula into something more pleasing to the eye? Irino. (talk) 05:49, 11 October 2013 (UTC)

According to the wikipedia page, the M25 is 117 miles long. That sounds more like "37 miles across" to me. Kaa-ching (talk) 08:46, 11 October 2013 (UTC)

Neither the sun or moon, nor Messier 25 (declination -19°) can ever culminate in the zenith over London. :-( Admittedly, Townsville, Australia would be sort of overwhelmed by M25. --129.13.72.198 11:27, 11 October 2013 (UTC)

M25 is a reference to the highway that surrounds London, not the Messier object, which is probably nowhere near the angular size of the moon. 65.129.214.100 15:17, 11 October 2013 (UTC)

Why would Randall choose London, if it wasn´t for the obvious disambiguity of the name M25? ----Wilberforce (talk) 12:33, 13 October 2013 (UTC)

Does anyone know why the exoplanet "HD 189733 b" is labled as "Permadeath" ? Same question for the other weird names in the same pannel (the "tilde on keyboard" one) ? Jahvascriptmaniac (talk) 11:32, 11 October 2013 (UTC)

If you were looking from the center of the earth, as the situation suggests, wouldn't the M25 be reversed, east-to-west, as you look at the sun and the moon?--76.105.133.220 16:09, 11 October 2013 (UTC)

I visualize it as looking down on Earth, with the "shadow" of the celestial object on top of the M25/soccer field/laptop/etc. 67.51.59.66 17:02, 11 October 2013 (UTC)

Yeah, I thought that too. It's a happy thought. Why, you ask? Well, with vastly diminished (or - in the course of time - zero) output from RTG power sources, they're like weakened (or inactivated) viruses - that we've sent out to the rest of the Universe, to any other intelligent lifeforms that may find them. What does that remind you of?

[PS - that's a happy thought because I choose to interpret from a cross-contamination standpoint. Which, in this case, allows them to observe us in our own locale, and establish our intrinsic nature - before a two-way interaction with us, in 'shared space' :P, and observing us through the medium of those interactions.] 220.224.246.97 18:46, 11 October 2013 (UTC)

Too bad he didn't do the Pleiades. I mean, instead of using the Vatican, he could have used something geeky: Bletchley Park or something (though that's probably not big enough). Homunq (talk) 14:11, 12 October 2013 (UTC)

Panel 6 (extrasolar planets)

My table doesn't really match the image. An earth sized Planet would be at some micrometer, their hosting stars are about some centimeter. Who is wrong? Me or Randall?--Dgbrt (talk) 19:49, 13 October 2013 (UTC)

Is it just me or is the laptop a MacBook Pro? Xyz (talk) 13:55, 14 October 2013 (UTC)

I'm a bit disappointed by the lack of extragalactical objects. Or did I miss something? Starblue (talk) 08:52, 17 October 2013 (UTC)

Moon shadow

If I understand correctly, the comic show the size of objects at the Earth surface. So, if the shadow of the Moon is projected on London, it will cover approximately all the space inside the M25 motorway ?If so, why it is said that a total solar eclipse will normally cover a band of about 250 km wide (and not 60 km wide) on Earth ? 24.200.202.45 09:38, 21 October 2013 (UTC)

The projection is to the center of the earth, not w.r.t. the sun like the shadow of a solar eclipse. Starblue (talk) 09:56, 21 October 2013 (UTC)

Look at the first picture at this comic and compare it this to this one (left): Solar_eclipse_types.svg. You are just behind the moon at the surface of the earth, and when the moon is not close enough a total eclipse will not happen (right). All distances and also angular sizes belong to the surface but not the center of the Earth.--Dgbrt (talk) 18:54, 21 October 2013 (UTC)

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